= Solution
For a <Killing vector field> $\xi$, set $K^{\mu\nu}=\nabla^\mu\xi^\nu=-K^{\nu\mu}$. Choose surface orientation conventions so that the <Komar charge>, up to overall normalization, is
$$
Q_\xi(V)=\tfrac12\int_{\partial V}K^{\mu\nu}\,dS_{\mu\nu}=\int_VJ_\xi^\mu\,dS_\mu,\qquad J_\xi^\mu=\nabla_\nu K^{\mu\nu}.
$$
This is the <Komar current from trace-reversed stress-energy> construction. The equality is the spacetime <Stokes theorem>. With $[\nabla_\mu,\nabla_\nu]V^\rho=R^\rho{}_{\sigma\mu\nu}V^\sigma$, the <Killing equation> and $\nabla_\nu\xi^\nu=0$ give $J_\xi^\mu=R^\mu{}_\nu\xi^\nu$. In four dimensions, at zero cosmological constant, the <Einstein field equations> give \b[the current]
$$
\boxed{J_\xi^\mu=8\pi G\left(T^\mu{}_\nu-\tfrac12T\delta^\mu{}_\nu\right)\xi^\nu.}
$$
Reversing orientation changes the common overall sign, not conservation. The <contracted Bianchi identity> gives
$$
\nabla_\mu J_\xi^\mu=\tfrac12\xi^\nu\partial_\nu R+R^{\mu\nu}\nabla_\mu\xi_\nu=0.
$$
An isometry preserves <scalar curvature>, and the second term contracts symmetric and antisymmetric tensors. Equivalently <stress-energy conservation>, the <Killing equation>, and $\mathcal L_\xi T=0$ cancel the matter expression; its trace derivative needs this last observation. \b[Hence $\boxed{\nabla_\mu J_\xi^\mu=0}$.] Nonzero cosmological constant adds $\Lambda\xi^\mu$ to the geometric current, also divergence-free.
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